3.447 $$\int x^2 \coth (a+b x) \text{csch}^2(a+b x) \, dx$$

Optimal. Leaf size=42 $-\frac{x \coth (a+b x)}{b^2}+\frac{\log (\sinh (a+b x))}{b^3}-\frac{x^2 \text{csch}^2(a+b x)}{2 b}$

[Out]

-((x*Coth[a + b*x])/b^2) - (x^2*Csch[a + b*x]^2)/(2*b) + Log[Sinh[a + b*x]]/b^3

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Rubi [A]  time = 0.0581759, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {5419, 4184, 3475} $-\frac{x \coth (a+b x)}{b^2}+\frac{\log (\sinh (a+b x))}{b^3}-\frac{x^2 \text{csch}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Coth[a + b*x]*Csch[a + b*x]^2,x]

[Out]

-((x*Coth[a + b*x])/b^2) - (x^2*Csch[a + b*x]^2)/(2*b) + Log[Sinh[a + b*x]]/b^3

Rule 5419

Int[Coth[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Csch[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csch[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^2 \coth (a+b x) \text{csch}^2(a+b x) \, dx &=-\frac{x^2 \text{csch}^2(a+b x)}{2 b}+\frac{\int x \text{csch}^2(a+b x) \, dx}{b}\\ &=-\frac{x \coth (a+b x)}{b^2}-\frac{x^2 \text{csch}^2(a+b x)}{2 b}+\frac{\int \coth (a+b x) \, dx}{b^2}\\ &=-\frac{x \coth (a+b x)}{b^2}-\frac{x^2 \text{csch}^2(a+b x)}{2 b}+\frac{\log (\sinh (a+b x))}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.114799, size = 55, normalized size = 1.31 $-\frac{x \coth (a)}{b^2}+\frac{\log (\sinh (a+b x))}{b^3}+\frac{x \text{csch}(a) \sinh (b x) \text{csch}(a+b x)}{b^2}-\frac{x^2 \text{csch}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Coth[a + b*x]*Csch[a + b*x]^2,x]

[Out]

-((x*Coth[a])/b^2) - (x^2*Csch[a + b*x]^2)/(2*b) + Log[Sinh[a + b*x]]/b^3 + (x*Csch[a]*Csch[a + b*x]*Sinh[b*x]
)/b^2

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Maple [A]  time = 0.03, size = 72, normalized size = 1.7 \begin{align*} -2\,{\frac{x}{{b}^{2}}}-2\,{\frac{a}{{b}^{3}}}-2\,{\frac{x \left ( bx{{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{2\,bx+2\,a}}-1 \right ) }{{b}^{2} \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}+{\frac{\ln \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(b*x+a)*csch(b*x+a)^3,x)

[Out]

-2*x/b^2-2/b^3*a-2*x*(b*x*exp(2*b*x+2*a)+exp(2*b*x+2*a)-1)/b^2/(exp(2*b*x+2*a)-1)^2+1/b^3*ln(exp(2*b*x+2*a)-1)

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Maxima [B]  time = 1.52887, size = 144, normalized size = 3.43 \begin{align*} -\frac{2 \,{\left ({\left (b x^{2} e^{\left (2 \, a\right )} - x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + x e^{\left (4 \, b x + 4 \, a\right )}\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac{\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{3}} + \frac{\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

-2*((b*x^2*e^(2*a) - x*e^(2*a))*e^(2*b*x) + x*e^(4*b*x + 4*a))/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) +
b^2) + log((e^(b*x + a) + 1)*e^(-a))/b^3 + log((e^(b*x + a) - 1)*e^(-a))/b^3

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Fricas [B]  time = 2.46102, size = 979, normalized size = 23.31 \begin{align*} -\frac{2 \, b x \cosh \left (b x + a\right )^{4} + 8 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 2 \, b x \sinh \left (b x + a\right )^{4} + 2 \,{\left (b^{2} x^{2} - b x\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b^{2} x^{2} + 6 \, b x \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right )^{2} -{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 4 \,{\left (2 \, b x \cosh \left (b x + a\right )^{3} +{\left (b^{2} x^{2} - b x\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{b^{3} \cosh \left (b x + a\right )^{4} + 4 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b^{3} \sinh \left (b x + a\right )^{4} - 2 \, b^{3} \cosh \left (b x + a\right )^{2} + b^{3} + 2 \,{\left (3 \, b^{3} \cosh \left (b x + a\right )^{2} - b^{3}\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b^{3} \cosh \left (b x + a\right )^{3} - b^{3} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-(2*b*x*cosh(b*x + a)^4 + 8*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + 2*b*x*sinh(b*x + a)^4 + 2*(b^2*x^2 - b*x)*cosh
(b*x + a)^2 + 2*(b^2*x^2 + 6*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*s
inh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x
+ a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*(2*b*x*co
sh(b*x + a)^3 + (b^2*x^2 - b*x)*cosh(b*x + a))*sinh(b*x + a))/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x + a)*sinh(
b*x + a)^3 + b^3*sinh(b*x + a)^4 - 2*b^3*cosh(b*x + a)^2 + b^3 + 2*(3*b^3*cosh(b*x + a)^2 - b^3)*sinh(b*x + a)
^2 + 4*(b^3*cosh(b*x + a)^3 - b^3*cosh(b*x + a))*sinh(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cosh(b*x+a)*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.17487, size = 188, normalized size = 4.48 \begin{align*} -\frac{2 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b x e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2 \, e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) - \log \left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}{b^{3} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{3} e^{\left (2 \, b x + 2 \, a\right )} + b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="giac")

[Out]

-(2*b^2*x^2*e^(2*b*x + 2*a) + 2*b*x*e^(4*b*x + 4*a) - 2*b*x*e^(2*b*x + 2*a) - e^(4*b*x + 4*a)*log(e^(2*b*x + 2
*a) - 1) + 2*e^(2*b*x + 2*a)*log(e^(2*b*x + 2*a) - 1) - log(e^(2*b*x + 2*a) - 1))/(b^3*e^(4*b*x + 4*a) - 2*b^3
*e^(2*b*x + 2*a) + b^3)